Question

A punter drops a ball from rest vertically 1 meter down onto his foot. The ball leaves the foot with a speed of 27 m/s at an angle 52° above the horizontal. What is the impulse delivered by the foot? (Use 0.41 kg for the mass of the ball. Assume the punter faces the +x-direction and that the +y-direction is upward. Enter the impulse in kg · m/s and the direction in degrees counterclockwise from the +x-axis.)

Answer 1

This is an impulse-momentum problem. An "impulse" is defined as a "change in momentum." That means we have to find the momentum of the ball before it was kicked, and the momentum after it was kicked, and then calculate their difference.

Momentum = mass*velocity, so we have some velocities to figure out. Let's declare up and to the right as positive.

The ball fell a meter at 9.8 m/s^2.

d = 1/2*g*t^2

1 = 4.9*t^2

t^2 = 1/4.9

t = 0.451 sec

9.8 m/sec^2 * 0.451 sec = 4.424 m/sec downward = -4.424 m/sec j

The mass of the ball is 0.41 kg.

Initial Momentum of the ball (Pi) before kick is:

Pi = 0.41kg*(-4.424m/sec j)

= -1.813 N*sec j.

The problem then gets a little tricky when we think about the kick. Our initial momentum only has a vertical component, but when the ball is kicked, it has both a vertical component and a horizontal component as given by the 52º angle. Since momentum is a vector quantity, we can't discount the horizontal component.

It's launched with a velocity of 27 m/sec at an angle of 52º. That means that:

Vx = 27*cos(52)

Vy = 27*sin(52)

and thus the final momentum (Pf) is:

Pf_x = 0.41*27*cos(52) = 6.815 N*sec i

Pf_y = 0.41*27*sin(52) = 8.723 N*sec j

The change in momentum is Pf - Pi:

(6.815 i + 8.723 j) - (-1.813 j) = 4.23 i + 10.536 j

Now do a vector sum:

?(6.815² + 10.536²) = 12.547 N*sec

To find the angle, recognize that tan = y/x, so arctan(y/x) = theta

arctan(10.536/6.815) = ~57.10 degrees

8.93 N*sec at an angle of 62 degrees.